-1

let's say $f: \mathbb{R}^n \to [0,+\infty]$ is a positive measurable function. I want to prove that $\int_{\mathbb{R^{n}}} f \, d\lambda=0$ if and only if $f(x)=0$ almost everywhere.

Let's say $f(x)=0$ almost everywhere. Name $g$ the zero function. Now it follows that $g$ and $f$ are almost everywhere the same. This means that the intergal of $f$ equals the integral of $g$ (which exists because we know that the integral of the zero function existst). This means that $\int_{\mathbb{R}^n} f \, d\lambda=\int_{\mathbb{R}^n} g \, d\lambda=0$

Now the other side. Let's say $\int_{\mathbb{R}^n} f \, d\lambda=0$ . I know I need to do something with $f$ is positive because when it's not this implication wouldn't work. I just don't know how to start with this side. How do you start to prove this or what are thet tools?

questmath
  • 875
  • 5
  • 11
  • 2
    Let $B:=f^{-1}(\mathbb R^+)$. Then $\int_{\mathbb R^n}f\mathrm d\lambda=\int_B f\mathrm d\lambda+\int_{\mathbb R^n\backslash B}f\mathrm d\lambda$. Can you show that the second term is $0$, while the first is $0$ or positive, depending on $\lambda(B)$? – Vercassivelaunos Nov 16 '20 at 19:03
  • The second term can't be negative because $f$ is positive, So it needs to be bigger or equal to zero. Supose it's bigger then zero then f(x) wouldn't be equal to zero almost everywhere? – questmath Nov 16 '20 at 19:06
  • 1
    @questmath Hint : A straigthforward way to prove this is to use Markov inequality (https://en.wikipedia.org/wiki/Markov%27s_inequality) – TheSilverDoe Nov 16 '20 at 19:07
  • @MichaelHardy I'm really sorry... I'm just trying my best to ask my question as good as possible...I'm sorry, I really didn't meant to do something wrong or be lazy. – questmath Nov 16 '20 at 19:24
  • 2
    @questmath don't worry about it. You're formatting wasn't ideal but it hardly matters since we can all understand what you wrote. – user293794 Nov 16 '20 at 19:27
  • 1
    @questmath There's nothing wrong with what you wrote, is just a question of typesetting –  Nov 16 '20 at 19:27
  • Alright, thank you. I will watch out for it the next time. – questmath Nov 16 '20 at 19:31
  • Use Chebychev's Inequality – user158796 Nov 17 '20 at 03:53

2 Answers2

2

Hint: suppose that $\lambda(\{x\mid f(x)>0\})>0$. Then argue that there exists $n$ so that $\lambda(\{x\mid f(x)>0)>1/n\})>0$. Now recall that the Lebesgue integral of $f$ is the supremum of the integral of all simple functions less than $f$. Can you find a simple function less than $f$ with positive integral to get a contradiction?

Michael Hardy
  • 1
user293794
  • 3,668
  • Maybe the function $g(x)=0$? – questmath Nov 16 '20 at 19:33
  • 1
    That function has integral 0 so it won't help. Try to draw a picture: you have that there is some set of positive measure (make it an interval for the purposes of drawing if you want) so that $f$ is bigger than $1/n$ on that interval. What does that tell you about the integral i.e. area under $f$? – user293794 Nov 16 '20 at 19:35
0

$$ \{ x : f(x)>0 \} = \{x: f(x)\ge 1\} \cup \bigcup_{n=1}^\infty \left\{x:\tfrac 1 {n+1} \le f(x) < \tfrac 1 n \right\} $$ If the set on the left has positive measure then one of the terms in this union has positive measure, and that measure times the corresponding lower bound on values of $f(x)$ in that set gives you a positive lower bound on the integral.

Michael Hardy
  • 1